Let the random variable $X$ be the number of independent Bernoulli trials needed to reach $r$ failures when the probability of success is p. Then $X$ follows a negative binomial distribution \begin{equation} P(X=x|r,p) = \begin{cases} \binom{x-1}{x-r}p^{x-r}(1-p)^{r} & \text{for $x \in \{1,2,3,\dots\}$}\\ 0 & \text{otherwise} \end{cases} \end{equation}
If now we consider $x,r \in \Bbb R^+$, the continous generalization of this negative binomial distribution seems to be \begin{equation} f(x;r,p) = \begin{cases} \frac{\Gamma(x)}{\Gamma(x-r+1)\Gamma(r)}p^{x-r}(1-p)^{r} & \text{for $x>0$}\\ 0 & \text{otherwise} \end{cases} \end{equation}
What is the expected value of $X$? With all the gamma functions in it, I have a hard time figuring out how to integrate this expression.
Thanks a lot for your help!
You have a problem with how you have specified the support with respect to the given parametrization. Specifically, $$\Pr[X = x \mid r, p] = \binom{x-1}{x-r} p^{x-r} (1-p)^r, \quad x \in \{1, 2, \ldots\}$$ is going to give you binomial coefficients with negative lower index for $x < r$, which means that for positive integer $r$, $$\Pr[X = x \mid r, p] = 0, \quad x \in \{1, \ldots, r-1\},$$ and for positive noninteger $r$, $\Pr[X = x \mid r, p]$ may be negative. Your generalization suffers from the same problem. Therefore, you have no reason to conclude that integrating the density will yield $1$. To illustrate, here is a plot of $$f(x;r,p) = \frac{\Gamma(x)}{\Gamma(x-r+1)\Gamma(r)} p^{x-r} (1-p)^r$$ for the choice $r = 5$, $p = 1/3$:
As you can see, the intervals on which $f$ are nonnegative are $x \in (0,1] \cup [2,3] \cup [4, \infty)$. This situation worsens as $r$ increases. It is actually a property inherited by the requirement that the discrete negative binomial PMF must have $0$ probability mass for $x < r$, combined with the fact that the gamma function is in a certain sense the unique analytic continuation of the factorial function.
Consequently, $f$ is not a density. Even if you correctly restrict the support by writing the negative binomial PMF as $$\Pr[X = x \mid r, p] = \binom{x-1}{x-r} p^{x-r}(1-p)^r, \quad x \in \{r, r + 1, r + 2, \ldots \},$$ this does not imply either of $$\int_{x=r-1}^\infty f(x;r,p) \, dx \quad \text{or} \quad \int_{x=r}^\infty f(x;r,p) \, dx$$ integrates to $1$, for reasons that should by now be obvious.